A Lipschitz Refinement of the Bebutov–Kakutani Dynamical Embedding Theorem

Abstract

We prove that an $\mathbb {R}$-action on a compact metric space embeds equivariantly in the space of one-Lipschitz functions $\mathbb {R}\rightarrow [0,1]$ if its fixed point set can be topologically embedded in the unit interval. This is a refinement of the classical Bebutov–Kakutani theorem (1968).

1 Introduction

The purpose of this short paper is to refine a classical theorem of Bebutov [2] and Kakutani [4] on dynamical systems. We call (X, T) a flow if X is a compact metric space and

$$\begin{aligned} T:\mathbb {R}\times X\rightarrow X, \quad (t, x) \mapsto T_t x \end{aligned}$$

is a continuous action of $\mathbb {R}$. We define $\mathrm {Fix}(X,T)$ (sometimes abbreviated to $\mathrm {Fix}(X)$) as the set of $x\in X$ satisfying $T_t x=x$ for all $t\in \mathbb {R}$. We define $C(\mathbb {R})$ as the space of continuous maps $\varphi : \mathbb {R}\rightarrow [0,1]$. It is endowed with the topology of uniform convergence over compact subsets of $\mathbb {R}$, namely the topology given by the distance

$$\begin{aligned} \sum _{n=1}^\infty 2^{-n} \max _{|t|\le n} |\varphi (t)-\psi (t)|, \quad (\varphi , \psi \in C(\mathbb {R})). \end{aligned}$$

(1.1)

The group $\mathbb {R}$ continuously acts on it by the translation:

$$\begin{aligned} \mathbb {R}\times C(\mathbb {R}) \rightarrow C(\mathbb {R}), \quad (s, \varphi (t))\mapsto \varphi (t+s). \end{aligned}$$

(1.2)

A continuous map $f:X\rightarrow C(\mathbb {R})$ is called an embedding of a flow (X, T) if f is an $\mathbb {R}$-equivariant topological embedding. Bebutov [2] and Kakutani [4] found that the $\mathbb {R}$-action on $C(\mathbb {R})$ has the following remarkable “universality”:

Theorem 1.1

(Bebutov–Kakutani) A flow (X, T) can be equivariantly embedded in $C(\mathbb {R})$ if and only if $\mathrm {Fix}(X,T)$ can be topologically embedded in the unit interval [0, 1].

The “only if” part is trivial because the set of fixed points of $C(\mathbb {R})$ is homeomorphic to [0, 1]. So the main statement is the “if” part.

Although the Bebutov–Kakutani theorem is clearly a nice theorem, it has one drawback: The space $C(\mathbb {R})$ is not compact (nor locally compact). So it is not a “flow” in the above definition. This poses the following problem:

Problem 1.2

Is there a compact invariant subset of $C(\mathbb {R})$ satisfying the same universality?

The purpose of this paper is to solve this problem affirmatively. Let $L(\mathbb {R})$ be the set of maps $\varphi :\mathbb {R}\rightarrow [0,1]$ satisfying the one-Lipschitz condition:

$$\begin{aligned} \forall s, t\in \mathbb {R}: \quad |\varphi (s)-\varphi (t)| \le |s-t|. \end{aligned}$$

(1.3)

$L(\mathbb {R})$ is a subset of $C(\mathbb {R})$. It is compact with respect to the distance (1.1) by Ascoli–Arzela’s theorem. The $\mathbb {R}$-action (1.2) preserves $L(\mathbb {R})$. So it becomes a flow. Our main result is the following. This solves [3, Question 4.1].

Theorem 1.3

A flow (X, T) can be equivariantly embedded in $L(\mathbb {R})$ if and only if $\mathrm {Fix}(X,T)$ can be topologically embedded in the unit interval [0, 1].

As in the case of the Bebutov–Kakutani theorem, the “only if” part is trivial because the fixed point set $\mathrm {Fix}(L(\mathbb {R}))$ is homeomorphic to [0, 1]. Since $L(\mathbb {R})$ is compact, it is a more reasonable choice of such a “universal flow”.

The proof of Theorem 1.3 is based on the techniques originally used in the proof of the Bebutov–Kakutani theorem (in particular, the idea of local section). A main new ingredient is the topological argument given in Sect. 2, which has some combinatorial flavor.

Remark 1.4

Problem 1.2 asks us to find a universal flow smaller than$C(\mathbb {R})$. If we look for a universal flow larger than$C(\mathbb {R})$, then it is much easier to find an example. Let $L^\infty (\mathbb {R})$ be the set of $L^\infty$-functions $\varphi : \mathbb {R}\rightarrow [0,1]$. (We identify two functions which are equal to each other almost everywhere.) We consider the weak$^*$ topology on it. Namely a sequence $\{\varphi _n\}$ in $L^\infty (\mathbb {R})$ converges to $\varphi \in L^\infty (\mathbb {R})$ if for every $L^1$-function $\psi :\mathbb {R}\rightarrow \mathbb {R}$

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{\mathbb {R}} \varphi _n(t) \psi (t) \, dt = \int _{\mathbb {R}} \varphi (t) \psi (t) \, dt. \end{aligned}$$

Then $L^\infty (\mathbb {R})$ is compact and metrizable by Banach–Alaoglu’s theorem and the separability of the space of $L^1$-functions, respectively. The group $\mathbb {R}$ acts continuously on it by translation. So it becomes a flow. Note that $\mathrm {Fix}\left( L^\infty (\mathbb {R})\right)$ is homeomorphic to [0, 1] and that the natural inclusion map $C(\mathbb {R}) \subset L^\infty (\mathbb {R})$ is an equivariant continuous injection. Then the Bebutov–Kakutani theorem implies the universality of $L^\infty (\mathbb {R})$: A flow (X, T) can be equivariantly embedded in $L^\infty (\mathbb {R})$ if and only if $\mathrm {Fix}(X,T)$ can be topologically embedded in [0, 1].

2 Topological Preparations

Let a be a positive number. We define L[0, a] as the space of maps $\varphi :[0,a]\rightarrow [0,1]$ satisfying

$$\begin{aligned} \forall s, t\in [0,a]: \quad |\varphi (s)-\varphi (t)| \le |s-t|. \end{aligned}$$

L[0, a] is endowed with the distance $\left| \left| \varphi -\psi \right| \right| _\infty = \max _{0 \le t\le a} |\varphi (t)-\psi (t)|$. We define $F_L[0, a] \subset L[0,a]$ as the space of constant functions $\varphi :[0,a]\rightarrow [0,1]$, which is homeomorphic to [0, 1].

Let (X, d) be a compact metric space. We define $C\left( X,L[0,a]\right)$ as the space of continuous maps $f:X\rightarrow L[0,a]$, which is endowed with the distance

$$\begin{aligned} \max _{x\in X} \left| \left| f(x)-g(x)\right| \right| _\infty . \end{aligned}$$

Lemma 2.1

Let $f\in C\left( X, L[0,a]\right)$ and suppose there exists $0<\tau <1$ satisfying

$$\begin{aligned} \forall x\in X, \forall s, t\in [0,a]: \quad |f(x)(s)-f(x)(t)| \le \tau |s-t|. \end{aligned}$$

(2.1)

Then for any $\delta >0$ there exists $g\in C\left( X, L[0,a]\right)$ satisfying

• (1) $\max _{x\in X} \left| \left| f(x)-g(x)\right| \right| _\infty <\delta$.

• (2) $g(x)(0)=f(x)(0)$ and $g(x)(a)=f(x)(a)$ for all $x\in X$.

• (3) $g(X)\cap F_L[0,a] = \emptyset$.

Proof

We take $0<b<c<a$ satisfying $b=a-c < \delta /4$. We take an open covering $\{U_1, \ldots ,U_M\}$ of X satisfying

$$\begin{aligned} \forall 1\le m\le M: \quad \mathrm {diam}f(U_m) < \min \left( \frac{\delta }{4}, \frac{(1-\tau )b}{2}\right) . \end{aligned}$$

(2.2)

We take a point $p_m\in U_m$ for each m. We choose a natural number N satisfying

$$\begin{aligned} N > M, \quad \Delta \overset{\mathrm {def}}{=} \frac{c-b}{N-1} < \frac{\delta }{4}. \end{aligned}$$

We divide the interval [b, c] into $(N-1)$ intervals of length $\Delta$:

$$\begin{aligned} b= a_1<a_2<\dots <a_{N}=c, \quad a_{n+1}-a_n = \Delta \quad (\forall 1\le n\le N-1). \end{aligned}$$

Set $A= \{a_1, \ldots , a_N\}$ and define a vector $e\in \mathbb {R}^A$ by $e=(1,1, \ldots , 1)$. Notice that $f(p_m)|_{A}$ is an element of $[0,1]^{A}$. Since $N>M$ we can choose $u_1, \ldots , u_M\in [0,1]^A$ satisfying

• (1) $|f(p_m)(a_n) - u_m(a_n)| < \min \left( \delta /4, (1-\tau )b/2\right)$ for all $1\le m\le M$ and $1\le n\le N$.

• (2) $|u_m(a_{n+1})-u_m(a_n)| < \Delta$ for all $1\le m\le M$ and $1\le n\le N-1$.

• (3) The $(M+1)$ vectors $e, u_1, \ldots , u_M$ are linearly independent.

Let $\{h_m\}_{m=1}^M$ be a partition of unity on X satisfying $\mathrm {supp}\, h_m\subset U_m$ for all m. For $x\in X$ we define a piecewise linear function $g(x):[0,a]\rightarrow [0,1]$ as follows. (We set $a_0=0$ and $a_{N+1}=a$.)

• $g(x)(0) = f(x)(0)$ and $g(x)(a)= f(x)(a)$.

• $g(x)(a_n) = \sum _{m=1}^M h_m(x) u_m(a_n)$ for $1\le n\le N$.

• We extend g(x) linearly. Namely, for $t= (1-\lambda ) a_n + \lambda a_{n+1}$ with $0\le \lambda \le 1$ and $0\le n\le N$ we set $g(x)(t) = (1-\lambda ) g(a_n) + \lambda g(a_{n+1})$.

$\square$

Claim 2.2

$g(x)\in L[0,a]$ and $\left| \left| g(x)-f(x)\right| \right| _\infty < \delta$.

Proof

For proving $g(x)\in L[0,a]$ it is enough to show $|g(x)(a_{n+1})-g(x)(a_n)| \le |a_{n+1}-a_n|$ for all $0\le n\le N$. For $1\le n\le N-1$, this is a direct consequence of the property (2) of $u_m$. So we consider the case of $n=0$. (The case of $n=N$ is the same).

$$\begin{aligned} \begin{aligned} |g(x)(b) - f(x)(0)| \le&\sum _{m=1}^M h_m(x) |u_m(b)-f(p_m)(b)| + \sum _{m=1}^M h_m(x) |f(p_m)(b) - f(x)(b)| \\&+ |f(x)(b)-f(x)(0)|. \end{aligned} \end{aligned}$$

We apply to each term of the right-hand side the property (1) of $u_m$, $\mathrm {diam}f(U_m) < (1-\tau )b/2$ in (2.2) and $|f(x)(b)-f(x)(0)| \le \tau b$ in (2.1) respectively. Then this is bounded by

$$\begin{aligned} \frac{(1-\tau )b}{2} + \frac{(1-\tau )b}{2} + \tau b = b. \end{aligned}$$

This proves $g(x)\in L[0,a]$.

Next we show $|g(x)(a_n) - f(x)(a_n)| < \delta /2$ for all $0\le n\le N+1$. For $n=0, N+1$, this is trivial. For $1\le n\le N$, we can bound $|g(x)(a_n)-f(x)(a_n)|$ from above by

$$\begin{aligned} \begin{aligned}&\sum _{m=1}^M h_m(x) |u_m(a_n)-f(p_m)(a_n)| + \sum _{m=1}^M h_m(x) |f(p_m)(a_n)-f(x)(a_n)| \\&\quad< \frac{\delta }{4}+ \frac{\delta }{4} = \frac{\delta }{2} \quad \left( \text {by the property (1) of}~u_m~\text {and}~\mathrm {diam}f(U_m) < \frac{\delta }{4}~\text {in}~(2.2)\right) . \end{aligned} \end{aligned}$$

Finally, let $a_n<t<a_{n+1}$. We can bound $|g(x)(t)-f(x)(t)|$ by

$$\begin{aligned} \begin{aligned}&|g(x)(t)-g(x)(a_n)| + |g(x)(a_n)-f(x)(a_n)| + |f(x)(a_n)-f(x)(t)| \\&\quad< 2(a_{n+1}-a_n) + \frac{\delta }{2} \quad (\text {by}~f(x), g(x)\in L[0,a]) \\&\quad< \delta \quad \left( \text {by}~a_{n+1}-a_n \le \max (b, \Delta ) < \frac{\delta }{4}\right) . \end{aligned} \end{aligned}$$

$\square$

For every $x\in X$, the function $g(x):[0,a]\rightarrow [0,1]$ is a non-constant function because

$$\begin{aligned} g(x)|_{\Lambda } = \sum _{m=1}^M h_m(x) u_m \not \in \mathbb {R} e \quad (\text {by the property (3) of}~u_m). \end{aligned}$$

This proves the statement. $\square$

We need two lemmas on linear algebra. For $u=(x_1, \ldots , x_{n+1})\in \mathbb {R}^{n+1}$ we set

$$\begin{aligned} Du = (x_2-x_1,x_3-x_2, \ldots , x_{n+1}-x_{n}) \in \mathbb {R}^{n}. \end{aligned}$$

Lemma 2.3

Let $l\ge m+1$ and set $e=(\underbrace{1,1, \ldots , 1}_l) \in \mathbb {R}^{l}$. The set of $(u_1, \ldots , u_m) \in \mathbb {R}^{l+1}\times \dots \times \mathbb {R}^{l+1} = \left( \mathbb {R}^{l+1}\right) ^m$ such that

$$\begin{aligned} \text {the vectors}~e, Du_1, Du_2, \ldots , Du_{m}~\text { are linearly independent} \end{aligned}$$

(2.3)

is open and dense in $\left( \mathbb {R}^{l+1}\right) ^{m}$.

Proof

The condition (2.3) defines a Zariski open set in $\left( \mathbb {R}^{l+1}\right) ^{m}$. So it is enough to show that the set is non-empty because a non-empty Zariski open set is always dense in the Euclidean topology. We set

$$\begin{aligned} u_i = (\underbrace{-1, \ldots , -1}_{i}, \, \underbrace{0, \ldots , 0}_{l+1-i}) , \quad (1\le i\le m). \end{aligned}$$

Then

$$\begin{aligned} Du_i = (\underbrace{0, \ldots , 0}_{i-1},\, 1, \, \underbrace{0, \ldots , 0}_{l-i}). \end{aligned}$$

The vectors $e, Du_1, \ldots Du_{m}$ are linearly independent. $\square$

Lemma 2.4

Let $n > l \ge 2m$. The set of $(u_1, \ldots , u_m) \in \mathbb {R}^n\times \dots \times \mathbb {R}^n = \left( \mathbb {R}^n\right) ^m$ such that, for any integer $\alpha$ with $2\le \alpha \le n-l+1$,

$$\begin{aligned} u_1|_1^l ,\> u_1|_{\alpha }^{\alpha +l-1} ,\> u_2|_1^l, \> u_2|_{\alpha }^{\alpha +l-1}, \ldots , u_m|_{1}^l, \> u_m|_{\alpha }^{\alpha +l-1}~ \text {are linearly independent in}~\mathbb {R}^l \end{aligned}$$

(2.4)

is open and dense in $\left( \mathbb {R}^n\right) ^m$. Here for $u_i = (x_{i1}, \ldots , x_{i n})$

$$\begin{aligned} u_i|_{1}^l = (x_{i 1}, \ldots , x_{i l}), \quad u_i|_\alpha ^{\alpha +l-1} = (x_{i, \alpha }, \ldots , x_{i, \alpha + l-1}). \end{aligned}$$

Proof

The defined set $A=\cap _{\alpha =2}^{n-l+1}A_\alpha$, where

$$\begin{aligned} A_\alpha =\{(u_1, \ldots ,u_m)\in \left( \mathbb {R}^n\right) ^m|\, (2.4)~\text {is satisfied}\}, \end{aligned}$$

is a Zariski open set in $\left( \mathbb {R}^n\right) ^m$. Hence it is enough to show that every $A_\alpha$ is nonempty. For each fixed $2 \le \alpha \le n-l+1$, we define $u_i = (x_{i1}, \ldots , x_{i n})$$(1 \le i\le m)$ by

$$\begin{aligned} x_{ij} = 1 \quad \left( j= i, \alpha + l-i \right) , \quad x_{ij} = 0 \quad (\text {otherwise}). \end{aligned}$$

Then it is direct to check that $(u_1, \ldots ,u_m)\in A_\alpha$. One can also use a proof from [5, Lemma 5.5]. $\square$

Lemma 2.5

Let $f\in C\left( X, L[0,a]\right)$ and suppose there exists $0<\tau <1$ satisfying (2.1). Then for any $\delta >0$ there exists $g\in C\left( X, L[0,a]\right)$ satisfying

• (1) $\max _{x\in X} \left| \left| f(x)-g(x)\right| \right| _\infty < \delta$.

• (2) $g(x)(0)=f(x)(0)$ and $g(x)(a)=f(x)(a)$ for all $x\in X$.

• (3) If $x,y\in X$ and $0\le \varepsilon \le a/2$ satisfy

  $$\begin{aligned} \forall t\in [0,a-\varepsilon ]: g(x)(t+\varepsilon ) = g(y)(t) \end{aligned}$$

  then $\varepsilon =0$ and $d(x,y) < \delta$.

Proof

Except for the use of the above two lemmas on linear algebra, the proof is close to Lemma 2.1. We take $0<b<c<a$ with $b=a-c < \min \left( \delta /4, a/4\right)$. We take an open covering $\{U_1, \ldots ,U_M\}$ satisfying $\mathrm {diam}\, U_m < \delta$ and $\mathrm {diam}f(U_m) < \min \left( \delta /4, (1-\tau )b/2\right)$ for all $1\le m\le M$. Take $p_m\in U_m$ for each m. Let $N\ge 2$ be a natural number and set $\Delta = (c-b)/(N-1)$. We introduce a partition $b=a_1<a_2<\dots <a_N=c$ by $a_n = b + (n-1)\Delta$. We set $A= \{a_1, \ldots , a_N\}$ and $\Lambda = A\cap [b,a/4] = \{a_1, \ldots , a_L\}$. We also set $e=(\underbrace{1,1, \ldots , 1}_L)\in \mathbb {R}^{L}$. We choose N sufficiently large so that

$$\begin{aligned} \Delta < \frac{\delta }{4}, \quad N> L \ge 2M. \end{aligned}$$

Since $L\ge 2M\ge M+1$, by using Lemmas 2.3 and 2.4, we can choose $u_1, \ldots , u_M\in [0,1]^A$ satisfying

• (1) $|f(p_m)(a_n) - u_m(a_n)| < \min \left( \delta /4, (1-\tau )b/2\right)$ for all $1\le m\le M$ and $1\le n\le N$.

• (2) $|u_m(a_{n+1})-u_m(a_n)| < \Delta$ for all $1\le m\le M$ and $1\le n\le N-1$.

• (3) Define $D_L u_m = (u_m(a_2)-u_m(a_1), \ldots , u_m(a_{L+1})-u_m(a_L))\in \mathbb {R}^L$. Then the $(M+1)$ vectors $e, D_L u_1, \ldots , D_L u_M$ in $\mathbb {R}^L$ are linearly independent.

• (4) For any $\varepsilon > 0$ with $\varepsilon + \Lambda \subset A$,

  $$\begin{aligned} u_1|_{\Lambda },\> u_1|_{\varepsilon +\Lambda },\> u_2|_{\Lambda },\> u_2|_{\varepsilon +\Lambda }, \ldots , u_m|_{\Lambda }, \> u_m|_{\varepsilon +\Lambda }~\text {are linearly independent in}~ \mathbb {R}^\Lambda \end{aligned}$$

For $x\in X$ we define $g(x):[0,a]\rightarrow [0,1]$ in the same way as in the proof of Lemma 2.1. Namely, we set $g(x)(0)= f(x)(0)$, $g(x)(a)= f(x)(a)$ and $g(x)(a_n) = \sum _{m=1}^M h_m(x) u_m(a_n)$ for $1\le n\le N$, where $\{h_m\}$ is a partition of unity satisfying $\mathrm {supp}\, h_m\subset U_m$. We extend g(x) to [0, a] by linearity. It follows that $g(x)\in L[0,a]$ and $\left| \left| g(x)-f(x)\right| \right| _\infty < \delta$ as before. We need to check the property (3) of the statement. Suppose there exist $x,y\in X$ and $0\le \varepsilon \le a/2$ satisfying $g(x)(t+\varepsilon ) = g(y)(t)$ for all $0\le t\le a-\varepsilon$.

First we show $\varepsilon + \Lambda \subset A$. Otherwise, $(\varepsilon +\Lambda ) \cap A = \emptyset$. Then it follows from the piecewise linearity that the function g(y)(t) becomes differentiable at every $t\in \Lambda$, which implies

$$\begin{aligned} g(y)(a_{n+1}) - g(y)(a_n) = g(y)(a_{n+2})-g(y)(a_{n+1}) \quad (1\le n\le L-1), \end{aligned}$$

and hence

$$\begin{aligned} \sum _{m=1}^M h_m(y) \left( u_m(a_{n+1}) - u_m(a_n)\right) = \sum _{m=1}^M h_m(y)\left( u_m(a_{n+2})-u_m(a_{n+1})\right) \quad (1\le n\le L-1). \end{aligned}$$

This means that $\sum _{m=1}^M h_m(y) D_L u_m \in \mathbb {R} e$, which contradicts the property (3) of $u_m$. So we must have $\varepsilon +\Lambda \subset A$.

The equation $g(x)(t+\varepsilon ) = g(y)(t)$$(0\le t\le a-\varepsilon )$ implies

$$\begin{aligned} \sum _{m=1}^M h_m(x) u_m|_{\varepsilon +\Lambda } = \sum _{m=1}^M h_m(y) u_m|_{\Lambda }. \end{aligned}$$

It follows from the property (4) of $u_m$ that $\varepsilon =0$ and $h_m(x)=h_m(y)$ for all $1\le m\le M$. Then $x,y\in U_m$ for some m and hence $d(x,y) \le \mathrm {diam}\, U_m < \delta$. $\square$

3 Proof of Theorem 1.3

Let (X, T) be a flow. Set $F=\mathrm {Fix}(X,T)$. We define $F_L = \mathrm {Fix}\left( L(\mathbb {R})\right)$. Namely $F_L$ is the space of constant maps $\varphi :\mathbb {R}\rightarrow [0,1]$, which is homeomorphic to [0, 1]. Suppose there exists a topological embedding $h:F\rightarrow F_L$. We would like to show that there exists an equivariant embedding $f:X\rightarrow L(\mathbb {R})$ with $f|_F=h$. We define $C_{T,h}\left( X,L(\mathbb {R})\right)$ as the space of equivariant continuous maps $f:X\rightarrow L(\mathbb {R})$ satisfying $f|_{F} = h$, which is endowed with the compact-open topology. For $f\in C_{T,h}\left( X, L(\mathbb {R})\right)$ we define $\mathrm {Lip}(f)$ as the supremum of

$$\begin{aligned} \frac{|f(x)(t)-f(x)(s)|}{|s-t|} \end{aligned}$$

over all $x\in X$ and $s,t\in \mathbb {R}$ with $s\ne t$.

Lemma 3.1

The space $C_{T,h}\left( X, L(\mathbb {R})\right)$ is not empty. Moreover for any $\delta >0$ there exists $f\in C_{T,h}\left( X, L(\mathbb {R})\right)$ satisfying $\mathrm {Lip}(f)\le \delta$.

Proof

Consider the map

$$\begin{aligned} F\ni x\rightarrow h(x)(0) \in [0,1]. \end{aligned}$$

By the Tietze extension theorem, we can extend this function to a continuous map $h_0:X\rightarrow [0,1]$. Let $\varphi : \mathbb {R}\rightarrow [0,1]$ be a smooth function satisfying

$$\begin{aligned} \int _{-\infty }^\infty \varphi (t)\, dt = 1, \quad \int _{-\infty }^\infty \varphi '(t) \, dt \le \min \left( 1, \delta \right) . \end{aligned}$$

For $x\in X$ we define $f(x): \mathbb {R} \rightarrow [0,1]$ by

$$\begin{aligned} f(x)(t) = \int _{-\infty }^\infty \varphi (t-s) h_0(T_s x) \, ds. \end{aligned}$$

Then $|f(x)'(t)|\le \min (1,\delta )$ and $f=h$ on F. Hence $f\in C_{T,h}\left( X, L(\mathbb {R})\right)$ and $\mathrm {Lip}(f)\le \delta$. $\square$

We borrow the next lemma from Auslander [1, p. 186, Corollary 6].

Lemma 3.2

Let $p\in X{\setminus } F$. There exist $a >0$ and a closed set $S\subset X$ containing p such that the map

$$\begin{aligned} {[}-a, a] \times S\rightarrow X, \quad (t, x)\mapsto T_t x \end{aligned}$$

(3.1)

is a continuous injection whose image contains an open neighborhood of p in X. We call (a, S) a local section around p and denote the image of (3.1) by $[-a,a] \cdot S$.

Proof

We explain the proof for the convenience of readers. We can find $c<0$ and a continuous function $h:X\rightarrow [0,1]$ satisfying $T_c p \not \in \mathrm {supp}\, h$ and $h=1$ on a neighborhood of p. We define $f:X\rightarrow \mathbb {R}$ by

$$\begin{aligned} f(x) = \int _c^0 h(T_t x) dt. \end{aligned}$$

We choose $0<a<|c|$ and a closed neighborhood A of p satisfying

$$\begin{aligned} \bigcup _{|t|\le a} T_t(A) \subset \{h=1\}, \quad \bigcup _{|t|\le a} T_{t+c}(A) \cap \mathrm {supp}\, h = \emptyset . \end{aligned}$$

It follows that $f\left( T_t x\right) = f(x) + t$ for any $x\in A$ and $|t|\le a$. Set $S= \{x\in A|\, f(x)=f(p)\}$. Then (a, S) becomes a local section. Indeed if $x,y\in S$ and $s,t\in [-a,a]$ satisfy $T_s x = T_t y$, then $s+f(p) = f(T_s x)= f(T_t y) = t+f(p)$ and hence $s=t$ and $x=y$. Thus the map (3.1) is injective. We take $0<b<a$ and an open neighborhood U of p satisfying $\bigcup _{|t|<b} T_t(U) \subset A$. Then the set $[-a,a]\cdot S$ contains

$$\begin{aligned} \{x\in U|\, -b<f(x)-f(p)<b\} \end{aligned}$$

(3.2)

because if $x\in U$ satisfies $t \overset{\mathrm {def}}{=} f(x)-f(p) \in (-b,b)$ then $f(T_{-t} x) = f(x)-t = f(p)$ (i.e. $T_{-t}x\in S$) and $x = T_t(T_{-t} x) \in [-a,a]\cdot S$. The set (3.2) is an open neighborhood of p. $\square$

Lemma 3.3

For any point $p\in X{\setminus } F$ there exists a closed neighborhood A of p in X such that the set

$$\begin{aligned} G(A) = \left\{ f\in C_{T,h}\left( X, L(\mathbb {R})\right) |\, f(A) \cap F_L = \emptyset \right\} \end{aligned}$$

(3.3)

is open and dense in the space $C_{T,h}\left( X, L(\mathbb {R})\right)$.

Proof

Take a local section (a, S) around p. For $x\in X$ we define $H(x)\subset \mathbb {R}$ (the set of hitting times) as the set of $t\in \mathbb {R}$ satisfying $T_t x\in S$. Any two distinct $s,t\in H(x)$ satisfy $|s-t|> a$. Notice that if $x\in F$ then $H(x)=\emptyset$. We denote by $\mathrm {Int}\left( [-a,a]\cdot S\right)$ the interior of $[-a,a]\cdot S$. We choose a closed neighborhood $A_0$ of p in S satisfying $A_0\subset \mathrm {Int}\left( [-a,a]\cdot S\right)$. We define a closed neighborhood A of p in X by

$$\begin{aligned} A = \bigcup _{|t|\le a} T_t(A_0). \end{aligned}$$

We choose a continuous function $q: S\rightarrow [0,1]$ satisfying $q=1$ on $A_0$ and $\mathrm {supp}\, q \subset \mathrm {Int}\left( [-a,a]\cdot S\right)$.

The set G(A) defined in (3.3) is obviously open. So it is enough to prove that it is dense. Take $f\in C_{T,h}\left( X, L(\mathbb {R})\right)$ and $0<\delta <1$. By Lemma 3.1 we can find $f_0\in C_{T,h}\left( X, L(\mathbb {R})\right)$ satisfying $\mathrm {Lip}(f_0)\le 1/2$. We define $f_1\in C_{T,h}\left( X, L(\mathbb {R})\right)$ by

$$\begin{aligned} f_1(x)(t) = (1-\delta ) f(x)(t) + \delta f_0(x)(t). \end{aligned}$$

It follows $\mathrm {Lip}(f_1) \le 1-\delta /2 < 1$. We apply Lemma 2.1 to the map

$$\begin{aligned} X\ni x\mapsto f_1(x)|_{[0,a]} \in L[0,a]. \end{aligned}$$

Then we find $g\in C\left( X, L[0,a]\right)$ satisfying

• (1) $|g(x)(t)-f_1(x)(t)| < \delta$ for all $x\in X$ and $0\le t\le a$.

• (2) $g(x)(0)= f_1(x)(0)$ and $g(x)(a)= f_1(x)(a)$ for all $x\in X$.

• (3) $g(X) \cap F_L[0,a] = \emptyset$.

We set $u(x)(t) = g(x)(t)-f_1(x)(t)$ for $x\in X$ and $0\le t\le a$. We define $g_1\in C_{T,h}\left( X, L(\mathbb {R})\right)$ as follows: Let $x\in X$.

• For each $s\in H(x)$, we set

  $$\begin{aligned} g_1(x)(t) = f_1(x)(t) + q(T_s x)\cdot u(T_s x)(t-s) \quad \text {for}~ t\in [s,s+a]. \end{aligned}$$

• For $t\in \mathbb {R}{\setminus } \bigcup _{s\in H(x)} [s,s+a]$, we set $g_1(x)(t) = f_1(x)(t)$.

This satisfies

$$\begin{aligned} |g_1(x)(t)-f(x)(t)| \le |g_1(x)(t)-f_1(x)(t)| + |f_1(x)(t)-f(x)(t)| \le 3\delta \end{aligned}$$

for all $x\in X$ and $t\in \mathbb {R}$. If $x\in A$ then there exists $s\in [-a,a]$ with $T_s x\in A_0$ and hence

$$\begin{aligned} g_1(x)(s+t) = g(T_s x)(t) \quad \text {for}~t\in [0,a]. \end{aligned}$$

It follows from the property (3) of g that the function $g_1(x)$ is not constant. Thus $g_1\in G(A)$. Since f and $\delta$ are arbitrary, this proves that G(A) is dense in $C_{T,h}\left( X, L(\mathbb {R})\right)$. $\square$

Lemma 3.4

For any two distinct points p and q in $X{\setminus } F$ there exist closed neighborhoods B and C of p and q in X respectively such that the set

$$\begin{aligned} G(B,C) = \left\{ f\in C_{T,h}\left( X, L(\mathbb {R})\right) |\, f(B) \cap f(C) = \emptyset \right\} \end{aligned}$$

(3.4)

is open and dense in $C_{T,h}\left( X, L(\mathbb {R})\right)$.

Proof

Take local sections $(a, S_1)$ and $(a, S_2)$ around p and q respectively. We can assume that $[-a,a]\cdot S_1$ and $[-a,a]\cdot S_2$ are disjoint with each other. For $x\in X$ we define H(x) as the set of $t\in \mathbb {R}$ satisfying $T_t x\in S_1\cup S_2$. We choose closed neighborhoods $B_0$ of p in $S_1$ and $C_0$ of q in $S_2$ respectively satisfying $B_0\subset \mathrm {Int}\left( [-a,a]\cdot S_1\right)$ and $C_0\subset \mathrm {Int}\left( [-a,a]\cdot S_2\right)$. We take a continuous function $\tilde{q}:X\rightarrow [0,1]$ satisfying $\tilde{q}=1$ on $B_0\cup C_0$ and $\mathrm {supp}\, \tilde{q}\subset \mathrm {Int}\left( [-a,a]\cdot S_1\right) \cup \mathrm {Int}\left( [-a,a]\cdot S_2\right)$. We define closed neighborhoods B and C of p and q respectively by

$$\begin{aligned} B = \bigcup _{|t|\le a/4} T_t(B_0), \quad C = \bigcup _{|t|\le a/4} T_t(C_0). \end{aligned}$$

The set G(B, C) defined in (3.4) is open. We show that it is dense. Take $f\in C_{T,h}\left( X, L(\mathbb {R})\right)$ and $0<\delta <1$. We can assume that

$$\begin{aligned} \delta < d(B_0,C_0) \overset{\mathrm {def}}{=} \min _{x\in B_0, y\in C_0} d(x,y). \end{aligned}$$

(3.5)

We define $f_1\in C_{T,h}\left( X, L(\mathbb {R})\right)$ exactly in the same way as in the proof of Lemma 3.3. It satisfies $\mathrm {Lip}(f_1) \le 1-\delta /2$ and $|f(x)(t)-f_1(x)(t)| \le 2\delta$ for all $x\in X$ and $t\in \mathbb {R}$.

We apply Lemma 2.5 to the map

$$\begin{aligned} X\ni x\mapsto f_1(x)|_{[0,a]} \in L[0,a]. \end{aligned}$$

Then we find $g\in C\left( X, L[0,a]\right)$ satisfying

• (1) $|g(x)(t)- f_1(x)(t)| < \delta$ for all $x\in X$ and $0\le t \le a$.

• (2) $g(x)(0)=f_1(x)(0)$ and $g(x)(a)=f_1(x)(a)$ for all $x\in X$.

• (3) If $x,y\in X$ and $0\le \varepsilon \le a/2$ satisfy

  $$\begin{aligned} \forall t\in [0,a-\varepsilon ]: g(x)(t+\varepsilon ) = g(y)(t) \end{aligned}$$

  then $d(x,y) < \delta$.

We set $u(x)(t) = g(x)(t)-f_1(x)(t)$ for $x\in X$ and $0\le t\le a$. We define $g_1\in C_{T,h}\left( X, L(\mathbb {R})\right)$ as before. Namely, for $x\in X$,

• For each $s\in H(x)$, we set

  $$\begin{aligned} g_1(x)(t) = f_1(x)(t) + \tilde{q}(T_s x)\cdot u(T_s x)(t-s) \quad \text {for}~t\in [s,s+a]. \end{aligned}$$

• For $t\in \mathbb {R}{\setminus } \bigcup _{s\in H(x)} [s,s+a]$, we set $g_1(x)(t) = f_1(x)(t)$.

This satisfies $|g_1(x)(t)-f(x)(t)| \le |g_1(x)(t)-f_1(x)(t)| + |f_1(x)(t)-f(x)(t)| \le 3\delta$.

We would like to show $g_1(B)\cap g_1(C) = \emptyset$. Suppose $x\in B$ and $y\in C$ satisfy $g_1(x)=g_1(y)$. There exist $|s_1|\le a/4$ and $|s_2|\le a/4$ satisfying $T_{s_1} x\in B_0$ and $T_{s_2} y\in C_0$. We can assume $s_1\le s_2$ without loss of generality. Set $\varepsilon =s_2-s_1 \in [0,a/2]$. We have

$$\begin{aligned} g_1(x)(s_1+t) = g(T_{s_1} x)(t) \text { and } g_1(y)(s_2+t) = g(T_{s_2} y)(t) \quad \text {for}~t\in [0,a]. \end{aligned}$$

$g_1(x)=g_1(y)$ implies that

$$\begin{aligned} g(T_{s_1} x) (t+\varepsilon ) = g(T_{s_2} y) (t) \quad \text {for}~t\in [0, a-\varepsilon ]. \end{aligned}$$

It follows from the property (3) of g that $d(T_{s_1}x, T_{s_2}y) < \delta$. Since $\delta < d(B_0, C_0) \le d(T_{s_1}x, T_{s_2}y)$, this is a contradiction. Therefore $g_1(B)\cap g_1(C) = \emptyset$. This proves the lemma. $\square$

Now we can prove Theorem 1.3. Note that X and $X\times X$ are hereditarily Lindelöf (that means that every open cover of a subspace has a countable subcover). Using these facts and applying Lemma 3.3 to each point in $X{\setminus } F$ and Lemma 3.4 to every pair of distinct points in $X{\setminus } F$, there exist countable families of closed sets $\{A_n\}_{n=1}^\infty$, $\{B_n\}_{n=1}^\infty$ and $\{C_n\}_{n=1}^\infty$ of $X{\setminus } F$ such that

• $X{\setminus } F = \bigcup _{n=1}^\infty A_n$ and $(X{\setminus } F)\times (X{\setminus } F) {\setminus }\{(x,x):x\in X\} = \bigcup _{n=1}^\infty B_n\times C_n$.

• $G(A_n)$ are open and dense in the space $C_{T,h}\left( X, L(\mathbb {R})\right)$ for all $n\ge 1$.

• $G(B_n,C_n)$ are open and dense in the space $C_{T,h}\left( X, L(\mathbb {R})\right)$ for all $n\ge 1$.

By the Baire category theorem, the set

$$\begin{aligned} \bigcap _{n=1}^\infty G(A_n) \cap \bigcap _{n=1}^\infty G(B_n,C_n) \end{aligned}$$

is dense and $G_\delta$ in $C_{T,h}\left( X, L(\mathbb {R})\right)$. In particular it is not empty. Any element f in this set gives an embedding of the flow (X, T) in $L(\mathbb {R})$.

Remark 3.5

The proof of the Bebutov–Kakutani theorem in [1, 4] used the idea of “constructing large derivative”. It is possible to prove Theorem 1.3 by adapting this idea to the setting of one-Lipschitz functions. But this approach seems a bit tricky and less flexible than the proof given above. The above proof possibly has a wider applicability to different situations (e.g. other function spaces).